# Tag Info

2

Your operator is $$Tf(x) = \int^\infty_{-\infty}1_{y\leq x}e^{-(x-y)}f(y)\,dy$$ which by change of variables $z=x-y$ becomes $$Tf(x) = \int^\infty_{-\infty}1_{z\geq 0}e^{-z} f(x-z)\,dz.$$ This is just the convolution operator $$f\mapsto \phi*f$$ with $$\phi(z) = 1_{z\geq 0}e^{-z}.$$ Now one of the fundamental inequalities (not hard to prove) about ...

2

$Tf=\phi*f$, where $\phi(t)=e^t\chi_{(-\infty,0)}(t)$. So $$\widehat{Tf}=\hat\phi\hat f.$$You can easily calculate $\hat\phi$; now the norm of $T$ is $||\hat\phi||_\infty$ and the spectrum of $T$ is the essential range of $\hat\phi$. (In other words, the supremum and the range, since $\hat\phi$ is continuous.) (Since $T$ is convolution with an $L^1$ ...

1

The intersection of a countable dense subset $M$ with the subspace $Y$ can be empty. A possibility would be to project $M$ onto $Y$ (i.e., choosing $p(x)\in Y$ such that $\|p(x)-x\| \le \|y-x\|$ for all $y\in Y$) and then show that $\lbrace p(x):x\in M\rbrace$ is dense.

1

The definition of $y$ is that $y\in G$ and $(x-y)\perp G$. Therefore, if $y' \in G$ $$x-y' = (x-y)+(y-y')$$ and the decomposition on the right is orthogonal because $y-y' \in G$. So $$\|x-y'\|^{2}=\|x-y\|^{2}+\|y-y'\|^{2} \ge \|x-y\|^{2},$$ with equality iff $y=y'$.

1

Let's state something more general first. The uniform boundedness principle implies that for any Banach space $X$, every weakly convergent sequence $\{x_n\}\subset X$ is bounded. Indeed, We have the canonical isometric embedding $j:X\to\ X^{**}$. The sequence $\{ j(x_n)y\}$ has a limit for every $y\in X^*$. Thus, the family of operators $\{j(x_n)\}$ is ...

6

A function with this property does not have to be in $L^2$. Let's first note the following lemma: Lemma. Let $H$ be a separable Hilbert space and $E \subset H$ a dense linear subspace. There exists a countable set $\{e_n\} \subset E$ which is an orthonormal basis for $H$. Proof of lemma. Since $H$ is separable, so is $E$. Let $\{x_n\}$ be a ...

0

Notice that for symmetric matrices $A$ and $B$, eigenvalue of $A\otimes B$ coincide with multiplication of eigenvalue of $A$ and $B$. For this you can see page 708 of Abstract harmonic analysis Hewitt&Ross(same method). About your question, Suppose $x\otimes w$ is an eigenvector of $A\otimes I$(thus $x$&$w$ arenot zero), $$(A\otimes I)(x\otimes ... 1 Observe that F^\perp = \ker P because P is an orthogonal projection and F = {\rm im} P. If PT = TP, then we have T(F) = T(P(H)) = P(T(H)) \subseteq P(H) = F so F is T-invariant. Similarly, P(T(F^\perp)) = T(P(F^\perp)) = T(P(\ker P)) = 0, hence T(F^\perp) \subseteq \ker P = F^\perp and F^\perp is T-invariant. Conversely, if F and ... 0 For two vectors x,y\in H, the notation x\otimes y is frequently used to denote the rank-one operator z\longmapsto \langle z,y\rangle x. It is related to tensor products, but not in the obvious way: it is seeing x\otimes y as an element of H\otimes H^* (most often, no one pays attention to this and just uses the notation). I think it is just a ... 1 You don't even need to know that every finite-dimensional subspace is closed. By Gram-Schmidt, given a countable Hamel basis, you can get a countable Hamel basis \{e_n\} that is orthonormal. But now if (a_n)\in \ell^2, the sequence of partial sums of \sum a_n e_n is Cauchy and hence converges to some element x. It is then easy to show that ... 1 Here is a complement to David C. Ullrich's counterexample. Theorem. Let \mu\in\mathcal{M}(\mathbb{R}^{d}) be a finite Borel measure. If \sum\left|\langle{\mu,\varphi_{j}}\rangle\right|^{2}<\infty for a complete orthonormal basis for L^{2}(\mathbb{R}^{d}), such that \text{span}\left\{\varphi_{j}\right\} is dense in ... 2 If I just gave the answer this post would be too short, heh-heh. A little discussion of how I came to the answer, possibly of interest: Decided I had no idea how to prove it, if true. Didn't see why it should be true, which proves nothing, since I have no feeling for an arbitrary orthonormal set. So I look for a counterexample. I noticed it is true for the ... 1 The zero operator has a nice basis of eigenvectors, all with the same eigenvalue. 0 The assertion is wrong: Given the interval [-1,1]. Consider the Borel measure:$$\mu:\mathcal{B}([-1,1])\to\overline{\mathbb{R}}:\quad\mu:=\lambda+\delta$$Regard the Borel measure:$$\rho:=1-1_{\{0\}}:\quad\mu_\rho(A):=\int_{[-1,1]}\rho\,\mathrm{d}\mu=\lambda$$It has Borel support:$$\sigma_\rho=\sigma(\lambda)=[-1,1]$$But it is not surjective: ... 1 The operator$$ Tf = f'-\frac{x}{\sqrt{1+x^{2}}}f $$is a bounded operator from X to Y because$$ \begin{align} |Tf| & \le |f'|+|f|,\\ |Tf|^{2} & \le |f'|^{2}+|f|^{2}+2|f'||f| \\ & \le 2|f'|^{2}+2|f|^{2} \\ \|Tf\|_{Y}^{2} & \le 2\|f\|_{X}^{2}. \end{align} $$To ... 0 This will fail even in finite dimensions, and even without an inner product. For example, in \mathbb{R}^4, consider the operator$$ A = \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}~. $$A is in Jordan form, and has eigenvalues 0 and 1, so in particular it ... 1 The implication is true. Use polarization. Suppose |a\eta+b\vartheta| is measurable for all complex a,b. Taking a=1, b=0 we see that |\eta| is measurable, and likewise so is |\vartheta|. Next, taking a=b=1, we see that$$|\eta + \vartheta|^2 = |\eta|^2 + |\vartheta|^2 + 2 \operatorname{Re}(\overline\eta \vartheta)$$is also measurable. So ... 1 As positive result: For orthogonal decompositions:$$\mathcal{H}=U\oplus V:\quad U\perp V\implies U=V^\perp,V=U^\perp$$Note it wasn't assumed closedness. 6 Not necessarily; this fails even in \mathbb{R}^2 with the usual inner product. Take U to be the x-axis, and V = \{(x,y):x=y\}, the diagonal. Then \mathbb{R}^2 = U\oplus V, but nothing other than 0 in V is orthogonal to anything in U other than 0. 2 Let k\ge n. Then by construction of s_k$$ \langle s_k,e_n\rangle = \langle \sum_{i=1}^k c_ie_i,e_n\rangle = c_k, where the latter equality follows from orthonormality of the e_n. Now c_n = \langle h,e_n\rangle , and the claim follows. 0 Assume that J=JJ^*J. If \varphi\in\mathcal R=(\ker J)^\perp=\overline{\mathcal R J^*}, then for any \xi\in\mathcal H we can write \xi=J^*\eta+\nu with \nu\in(\mathcal RJ^*)^\perp=\ker J, so \nu\perp\varphi and \begin{align} \langle J^*J\varphi,\xi\rangle=\langle J^*J\varphi,J^*\eta+\nu\rangle=\langle J^*J\varphi,J^*\eta\rangle=\langle ... 2 One argument is as follows: we can simply state that r(T) \leq r(T^*T) \leq r(T^*), so that r(T) = r(T^*). To prove that r(T) \leq r(T^*T), it suffices to prove that n(T^*T) \leq n(T). To prove this, note that T^*Tx = 0 \implies\\ (x,T^*Tx) = 0 \implies\\ (Tx,Tx) = 0 \implies\\ Tx = 0 $$so that \ker(T^*T) \subseteq \ker(T). 5 I'm not sure about the Laplace transform but in Joel L. Schiff's "The Laplace Transform: Theory and Applications" on page 13 the author proves that a large class of functions has a Laplace transform. I am not sure how to describe nicely the result in terms of domain and range of the operator, buy maybe that helps. As for the Fourier transform, you first ... 1 This should not be true because every term in the series has the same compact support. Since this is a solution to the heat equation, it should spread out. Edit: Another way of looking at this follows from the "uncertainty principle" : the Fourier transform of a "bump function" cannot decay exponentially fast. Thus we expect ||(\frac {d}{dx})^n f|| = ... 1 Hint: Since \|x_n\| is bounded, there is a subsequence (x_{n(k)})_{k\geq 1} that converges weakly to some x\in H. Show that this x is a solution. 4 This a potential answer to the question. It is not clear whether its proof technique can be completed or not. We can obtain the result by considering convergence of the Fourier transform of the partial sums. Since f is smooth and compactly supported it is in L^2(\mathbb{R}) as are all of its derivatives. Let f_n be the nth partial sum in the ... 0 You forgot another constraint that z has to follow which is z \in (\ker f)^\perp. Then it is true that if z_1 and z_2 are such that f(z_1)\neq 0 and f(z_2) \neq 0, \frac{z_1}{f(z_1)\|z_1\|^2}=\frac{z_2}{f(z_2)\|z_2\|^2} and z_1,z_2 \in (\ker f)^\perp you have z_1=z_2. 0 I was gonna shut up on this, but I can't stand it. Here's a much cleaner proof. Let X be the C^* algebra generated by (U_t). Basic C^* algebra stuff shows that X is isomorphic to C(K) for some compact Hausdorff space K. The isomorphism is an isometry and takes adjoints to complex conjugates. Say f_t\in C(K) corresponds to U_t. The fact ... 2 If t\in [0,\infty)\mapsto U(t)\in\mathcal{L}(H) is continuous in the operator norm, then, for \delta small enough, \frac{1}{\delta}\int_{0}^{\delta}U(t)dt is invertible because there exists \delta > 0 such that$$ \left\|\frac{1}{\delta}\int_{0}^{\delta}U(t)dt-I\right\|= \left\|\frac{1}{\delta}\int_{0}^{\delta}(U(t)-I)dt\right\| < 1. $$So ... 1 EDIT: "uniform operator topology" was just "operator topology" in the original, which I assumed meant either the strong or weak operator topology, in which case it's false: Say (X,\mu) is a measure space, suppose \alpha:X\to\Bbb R is measurable, and define U_t:L^2\to L^2 by$$U_tf(x)=e^{it\alpha(x)}f(x).$$Then (U_t) is strong-operator-continuous ... 1 Yes, your reasoning is correct (since closed operators are closable), so we can indeed include that$$ (\lambda I - T)^* = \overline \lambda I - T^* $$1 You can just use the definition. The adjoint of A is defined as the operator A^* such that \langle Ax,y\rangle = \langle x,A^* y\rangle. Therefore, for A=\lambda I - T,$$ \langle Ax,y\rangle=\langle (\lambda I-T)x,y\rangle = \lambda \langle x,y\rangle - \langle Tx,y\rangle = \langle x,\overline{\lambda}y\rangle - \langle x,T^*y\rangle = \langle x, ...

1

If $\rho$ is a bounded measurable function on $[a,b]$, then $E(t)=\int_{a}^{t}\rho(u)du$ is a function of bounded variation and, for any continuous function $g$, $$\int_{a}^{b}g(t)dE(t) = \int_{a}^{b}g(t)\frac{dE}{dt}dt = \int_{a}^{b}g(t)\rho(t)dt.$$ In your case, $$\|E(t)x\|^{2}=\int_{-\infty}^{t}|x(u)|^{2}du$$ The above easily ...

0

Finally I got it!! :D Reducibility It acts on the range: $$J^*J|A|=1_\overline{\mathcal{R}|A|}|A|=|A|$$ As it is bounded: $$J^*J\in\mathcal{B}(\mathcal{H},\mathcal{H}):\quad(J^*J|A|)^*=|A|J^*J$$ So one obtains: $$|A|J^*J=(J^*J|A|)^*=|A|^*=|A|$$ Concluding reducibility. Positivity Regard the selfadjoint: $$S:=J|A|J^*=(J|A|J^*)^*=S^*$$ Note that it is: ...

1

for first question see page 27 of Gohberg, I. C, Krein, M. G. Introduction to the Theoryof Linear Non-Self-Adjoint Operators in Hilbert Space. For $A\in\mathscr{I}_1(H)$ and $B\in\mathcal{B}(H).$we have $s_{j}(AB)\leq{s_{j}{(A)}} \Vert{B}\Vert_\infty$ & $s_{j}(BA)\leq{s_{j}{(A)}} \Vert{B}\Vert_\infty$.

1

It is correct for positive operators, for more information see Gohberg, Krein, Introduction to the Theoryof Linear Non-Self-Adjoint Operators in Hilbert Space page 27

4

Proof by contrapositive certainly works. If $a_{n} \not \to 0$, there is an $\epsilon > 0$ and subsequence $a_{n_k}$ such that $|a_{n_k}| > \epsilon$ for each $k$. Then, show that the sequence $\{e_{n_k}\}$ is mapped to a sequence with no convergent subsequence.

2

I finally got it.. :D By another thread: $$\|K\varphi\|=\langle K^*K\varphi,\varphi\rangle\leq\|K^*K\varphi\|\cdot\|\varphi\|$$ So equivalence holds: $$K\in\mathcal{K}(\mathcal{H})\iff K^*K\in\mathcal{K}(\mathcal{H})$$ Thus one has the chain: BR\in\mathcal{K}(\mathcal{H})\implies ... 2 That result quoted in the solution is not quite true: what is true is that \text{Ker } T = (\text{Im } T^*)^{\perp}. It doesn't matter, though, since T being compact implies that its Hilbert space adjoint T^* is compact as well. Hence, by Theorem 2, (\text{Im } T^*)^{\perp} \neq \{0\}, so that there is an e \neq 0 in H so that Te=0. 0 This, as you've stated, is not true: the function x \mapsto e^{-2\pi i x} is of course orthogonal to all the elements on the aforementioned set, and is nonzero. In order to obtain an orthonormal basis, you should've included \{ e^{-2\pi i k x}; k \in \Bbb{N} \} in the definition of the set above. To prove this set is really an orthonormal basis, you ... 0 I tried an other way: by composition and multiplication rules. Let's call g:=1/\|\bullet \|. Dg(x).(h)=-\frac 1 {||x||^3}<x,h> (because g=(1/\bullet) \circ f so  Dg(x)=-1/f(x)^2.Df(x) ) T=Id . g (so using D(fg)_{a}(h)=Df_{a}(h)g(a)+f(a)Dg_{a}(h)) we have DId(x)(h)=h (linear application get the same derivative). finally ... 3 Let \{ x_n \}_n  be a sequence, bounded say by M>0. If T^*T is compact, there exist a subsequence \{ T^*T(x_{n_k})\}_k that converges and hence is Cauchy. Let k<k' \in \mathbb{N}, then \begin{align*} \|T(x_{n_k}) - T(x_{n_{k'}}) \|^2 & =\| T(x_{n_k}-x_{n_{k'}}) \|^2 \\ & = \langle T(x_{n_k}-x_{n_{k'}}), T(x_{n_k}-x_{n_{k'}}) \rangle ... 2 So, you want to differentiate the map T(X) = x/\|x\| . For any vector h, the derivative of T(x+th) with respect to t at t=0 can be computed using the ordinary quotient rule:\frac{d}{dt}_{| t=0} \frac{x+th}{\|x+th\|} = \frac{h\|x\| - x\langle x,h\rangle/\|x\| }{\|x \|^2 }$$So, the derivative at x is the linear operator$$ h\mapsto ...

0

Independence of trace, is a result of $trace(AB)=trace(BA)$ & when you change basis, for new matrix $B$ there exist a unitary matrix like $P$ such that $A=PBP^{-1}$.

0

Let $U=PSQ^\ast$ be a singular value decomposition. Since $U$ has orthonormal rows, $S=(I_r,0_{r\times(n-r)})$. Therefore $UDU^\ast$ is unitarily equivalent to the leading principal $r\times r$ submatrix of $Q^\ast DQ$. In other words, you are essentially asking about the relationship between the eigenvalues of an arbitrary normal matrix and the eigenvalues ...

1

For clarity's sake, I've rewritten this second to last line as $$\left< x, x - \sum_{n=1}^{m}\left[ <x,e_{n}>e_{n}\right] \right> - \sum_{n=1}^{m}\left[ <x, e_{n}> \left < e_{n}, x - \sum_{i=1}^{m} <x,e_{i}>e_{i} \right >\right].$$ By linearity of the inner-product, $$\left< x, x - \sum_{n=1}^{m}\left[ ... 2 Note that the first inner product in the second-last line is:$$ \left\langle x, x - \sum_{n=1}^m \langle x,e_n\rangle e_n\right\rangle = \langle x,x\rangle - \sum_{n=1}^m \langle x,e_n\rangle \langle x,e_n\rangle = \|x\|^2 - \sum_{n=1}^m |\langle x,e_n\rangle|^2 $$so we would hope that the second is equal to zero. We have$$ \sum_{n=1}^m \langle ...

0

The eigenvalues of $M$ are convex combinations of the eigenvalues (i.e. the diagonal entries) of $D$. To see this, note that $P=U^*U$ is a projection, since $P^2=U^*UU^*U=U^*U=P$. Concretely, it is the projection onto the subspace spanned by the rows of $U$. Since the eigenvalues of $AB$ are the same as the eigenvalues of $BA$, the eigenvalues of ...

3

Because of the subspaces being selfadjoint, $B(H)V_1\subset V_2$ implies that $V_1B(H)\subset V_2$. If $V_1\ne0$, then $B(H)V_1B(H)\subset V_3$ contains all finite-rank operators, and thus $V_3$, being closed, contains the compact operators. If $V_3$ contains a non-compact operator, then the ideas in this answer show that $I\in V_7$ (I didn't count ...

0

Yeaaahh, I got it! :D Great thanks to David C. Ullrich!!! Counterexample Given the Hilbert space $\mathbb{C}^4$. Regard the matrices: $$N:=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\oplus\begin{pmatrix}0&0\\0&0\end{pmatrix}\quad N':=\begin{pmatrix}0&0\\0&0\end{pmatrix}\oplus\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ Then they are ...

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